According to the "Flat Earth Club", Earth is a plane described by the equation
x + y + z = 18:
Also according to the Flat Earth Club, Earth will be destroyed on the day this assignment
is due by an explosion that will spontaneously occur at the so-called \armageddon point"
A = (1; 1; 1). Point B will be the first point destroyed. Calculate Point B.

Sep 11th 2012, 07:47 AM

Prove It

Re: Simple Plane and Point problem.

Quote:

Originally Posted by aqualiary

According to the "Flat Earth Club", Earth is a plane described by the equation
x + y + z = 18:
Also according to the Flat Earth Club, Earth will be destroyed on the day this assignment
is due by an explosion that will spontaneously occur at the so-called \armageddon point"
A = (1; 1; 1). Point B will be the first point destroyed. Calculate Point B.

I assume that your explosion will take on the form of a sphere with centre at (1,1,1) and increasing radius. So you will be looking for the first point that

intersects with .

From the first equation we have

So the intersection with the plane will happen on the sphere centred at (0,0,0) of radius .

Sep 11th 2012, 09:11 AM

MaxJasper

Re: Simple Plane and Point problem.

Quote:

Originally Posted by aqualiary

According to the "Flat Earth Club", Earth is a plane described by the equation
x + y + z = 18:
Also according to the Flat Earth Club, Earth will be destroyed on the day this assignment
is due by an explosion that will spontaneously occur at the so-called \armageddon point"
A = (1; 1; 1). Point B will be the first point destroyed. Calculate Point B.

According to the "Flat Earth Club", Earth is a plane described by the equation
x + y + z = 18:
Also according to the Flat Earth Club, Earth will be destroyed on the day this assignment
is due by an explosion that will spontaneously occur at the so-called \armageddon point"
A = (1; 1; 1). Point B will be the first point destroyed. Calculate Point B.

The problem amounts to finding the point B in the plane that's closest to A in . I'll do it using vectors.

Although this setup might seem long and involved, it's actually, I think, by far the easiest way to do the problem. It just takes doing several such problems to get used to all the vector-stuff, but once you do, it makes everything seem almost trivial. I consider it a vastly more simple and direct than any method that calculates the distance from the point to the plane.

A plane P in (or any hyper plane in any ) is uniquely determined by a point on it, and a non-zero vector normal to it, via

, where is a non-zero normal vector, and is any vector in that plane.

For this problem, , where , and, say, .

(There, I chose using the coefficients (1, 1, 1) of the plane's equation x + y + z = 18, and by picking any point in that plane - I choose (18,0,0).)

Let , corresponding to A = (1, 1, 1). The vector that's closest to in is given by:

, where is the perpenticular line from to . But "direction ("slope") of " is parallel to , since .

Thus is given by , where is any vector in . Choose .

THUS THIS IS THE PROBLEM:
FIND , where , and ,

and where and .

SOLUTION:Have . Thus , so . Also, , so for some .

Finding will find . Substitute the second equation into the first to get and solve an equation for .

Get:

.

Now plug in the values , and to find :

Have

.

and .

Thus .

Therefore .

The answer to the problem is that the closest point B in that plane to the point A is B = (6, 6, 6).

(There's a math-religion-geek joke there that I didn't catch at first. What's the "armageddon point"? It's 666.)